答案：
1. B 点拨:如答图，所得的折线的函数表达式为
$y=\begin{cases}-2x+1(x＜\dfrac{1}{2}),\\2x-1(x≥\dfrac{1}{2}).\end{cases}$

当$x=-1$时,$y=-2×(-1)+1=3$,即$A(-1,3)$;
当$x=2$时,$y=4-1=3$,即$B(2,3)$;
当$y=0$时,$x=0.5$,即$C(0.5,0)$.
把$(-1,3)$代入$y=x+b$中,可得$b=4$;
把$(2,3)$代入$y=x+b$中,可得$b=1$;
把$(0.5,0)$代入$y=x+b$中,可得$b=-\dfrac{1}{2}$.
$\because$函数$y=|2x-1|$的图象与直线$y=x+b$的图象交点的横坐标$x$均满足$-1＜x＜2$,$\therefore\begin{cases}b＜4,\\b＜1,\\b≥-\dfrac{1}{2},\end{cases}$
即$-\dfrac{1}{2}≤ b＜1$.

答案：
2.$y=-\dfrac{1}{3}x+1$或$y=3x+1$ 点拨:$\because$一次函数$y=\dfrac{1}{2}x+1$的图象与$y$轴交于点$A$,与$x$轴交于点$B$,$\therefore A(0,1)$,$B(-2,0)$.
如答图①,当直线$y=\dfrac{1}{2}x+1$绕点$A$顺时针旋转$45°$后的图象为直线$l_1$,过点$B$作$BD⊥$直线$l_1$于点$D$,过点$D$作$DF⊥ y$轴于点$F$,过点$B$作$BE⊥ FD$交$FD$的延长线于点$E$,则$△ ABD$为等腰直角三角形,$\therefore AD=BD$.
$\because∠ FDA+∠ FAD=∠ FDA+∠ EDB=90°$,
$\therefore∠ FAD=∠ EDB$.
又$\because∠ AFD=∠ E=90°$,$\therefore△ ADF≌△ DBE(\mathrm{AAS})$,
$\therefore DE=AF$,$DF=BE$.
设$AF=a$,$\because A(0,1)$,$B(-2,0)$,
$\therefore DF=BE=OF=1+a$,$EF=ED+DF=a+1+a=OB=2$,
$\therefore a=\dfrac{1}{2}$,$\therefore DF=OF=1+a=\dfrac{3}{2}$,$\therefore D(-\dfrac{3}{2},\dfrac{3}{2})$.
设直线$l_1$的函数表达式为$y=kx+1$,则$\dfrac{3}{2}=-\dfrac{3}{2}k+1$,
解得$k=-\dfrac{1}{3}$,$\therefore$直线$l_1$的函数表达式为$y=-\dfrac{1}{3}x+1$.

如答图②,直线$y=\dfrac{1}{2}x+1$绕点$A$逆时针旋转$45°$后的图象为直线$l_2$,过点$B$作$BD⊥$直线$l_2$于点$D$,过点$D$作$DF⊥ y$轴于点$F$,作$DE⊥ x$轴于点$E$,则$△ ABD$为等腰直角三角形.
同理可证$△ ADF≌△ BDE$,$\therefore DF=DE$,$AF=BE$.
设$DF=b$,则$DE=b$.
$\because A(0,1)$,$B(-2,0)$,
$\therefore AF=BE=1+b$,$BO=BE+OE=b+1+b=2$,
$\therefore b=\dfrac{1}{2}$,$\therefore D(-\dfrac{1}{2},-\dfrac{1}{2})$.
设直线$l_2$的函数表达式为$y=kx+1$,
则$-\dfrac{1}{2}=-\dfrac{1}{2}k+1$,解得$k=3$,
$\therefore$直线$l_2$的函数表达式为$y=3x+1$.
综上可知,旋转后的图象对应的函数表达式是$y=-\dfrac{1}{3}x+1$或$y=3x+1$.

答案：
3.解:(1)过点$C$作$CE⊥ x$轴于点$E$,
$\because∠ ABC=90°$,
$\therefore∠ ABO+∠ CBE=90°$,
$\because∠ ABO+∠ BAO=90°$,$\therefore∠ CBE=∠ BAO$.
$\because AB=BC$,$\therefore△ ABO≌△ BCE(\mathrm{AAS})$,
$\therefore AO=BE=6$,$BO=CE=3$,$\therefore C(9,3)$.设直线$AC$的函数表达式为$y=kx+6$,
$\therefore 9k+6=3$,解得$k=-\dfrac{1}{3}$,$\therefore y=-\dfrac{1}{3}x+6$.
(2)点$B$关于点$H$的对称点$E(-3,2h)$,点$C$关于点$H$的对称点$F(-9,2h-3)$,
当点$E$在直线$AC$上时,$-\dfrac{1}{3}×(-3)+6=2h$,解得$h=\dfrac{7}{2}$;当点$F$在直线$AC$上时,$-\dfrac{1}{3}×(-9)+6=2h-3$,解得$h=6$,$\therefore\dfrac{7}{2}≤ h≤6$时,线段$EF$与直线$AC$有交点;
(3)$\because△ ABC$是等腰直角三角形,$\therefore∠ BAC=45°$,当点$B$落在直线$AC$上时,$AC'$与$AB$垂直,如答图,设直线$AC'$与$x$轴的交点为$M$,$\because BC⊥ AB$,$\therefore AM// BC$.
易知直线$BC$的函数表达式为$y=\dfrac{1}{2}x-\dfrac{3}{2}$,$\therefore$直线$AM$的函数表达式为$y=\dfrac{1}{2}x+6$.
设点$C'$的坐标为$(m,\dfrac{1}{2}m+6)$,
$\because A(0,6)$,$C(9,3)$,$\therefore AC=3\sqrt{10}$,
$\therefore AC'=3\sqrt{10}=\sqrt{m^2+\dfrac{1}{4}m^2}$,
解得$m=6\sqrt{2}$或$m=-6\sqrt{2}$,
$\therefore$旋转后的点$C$的对应点$C'$的坐标为$(6\sqrt{2},3\sqrt{2}+6)$或$(-6\sqrt{2},-3\sqrt{2}+6)$.